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Definition: The set of integers is $\mathbb{Z}:=\frac{\mathbb{N} \times \mathbb{N}}{R}=\{[(m,n)]:(m,n)\in \mathbb{N}\times \mathbb{N}\}$.

I understand this is the set of all the equivalence classes of $\mathbb{N}\times \mathbb{N}$ using the equivalence relation $R$, but who are $m$ and $n$? place holders for what?

Introducing, the relation of equivalence: $(m,n)R(p,q)\Leftrightarrow m + q = n + p.$

I didn't see this before!

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  • $\begingroup$ Also, can there be given an explicit relation for this? $\endgroup$ Commented May 25, 2016 at 23:33
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    $\begingroup$ You should state the definition of the equivalence relation. to make the question specific. [Wherever you found this definition should have that definition in it.] $\endgroup$
    – coffeemath
    Commented May 25, 2016 at 23:34
  • $\begingroup$ Done, i was missing a paragraph. $\endgroup$ Commented May 25, 2016 at 23:39
  • $\begingroup$ One way of seeing it is that after modding out by that relation, $(m,n)$ "represents" $m-n$, in the sense that each pair with the same value of $m-n$ is in the same class. (Of course this is a bit handwavy because we don't have the negative integers yet, but still, this is the idea.) $\endgroup$
    – Ian
    Commented May 25, 2016 at 23:42
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    $\begingroup$ @coffeemath The last one should be $(0,|z|)$ or $(0,-z)$. $\endgroup$
    – Ian
    Commented May 25, 2016 at 23:45

2 Answers 2

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Well, what do we think of when we think of the integers? The integers are the natural numbers, $0$, and the negative natural numbers. In the natural numbers, we have addition, but by introducing the integers, we get subtraction. Therefore, we can say that any integer is just the difference of two natural numbers: For any $z \in \Bbb{Z}$, there are $m, n \in \Bbb{N}$ such that $m-n=z$.

Now that we know that, how can we use that to form a formal definition of the integers? Well, we know we need two natural numbers $m, n$. We can get this from the set $\Bbb{N} \times \Bbb{N}$, which is the set of ordered pairs of natural numbers. Then, to get an integer from the ordered pair, we just map $(m, n)$ to $m-n$.

However, if we have two ordered pairs $(m, n)$ and $(p, q)$ where $m-n=p-q$, then we have two ordered pairs that map to the same integer. Therefore, we need these two ordered pairs to be "equal" somehow. This is where $R$ comes in. $(m, n)R(p, q)$ is the same as $m-n=p-q$. However, we want to base $R$ solely off the natural numbers, meaning no subtraction. Therefore, we manipulate the equation to get: $$(m, n)R(p, q) \iff m+q=p+n$$


Thus, that's how these people got this definition of the integers: They partitioned ordered pairs of natural numbers based off the difference of their elements. For example, here's the set of ordered pairs that represent $0$: $$0=\{(0, 0), (1, 1), (2, 2), ...\}$$ All of these elements have a difference of $0$ between their elements. If you test any of them against the $(m, n)R(p, q)$ relation, the relation holds true: $$(1, 1)R(2, 2) \iff 1+2=1+2 \iff 3=3$$

Now, here's another example: $$2=\{(2, 0), (3, 1), (4, 2), ...\}$$ All of these elements have a difference of $2$ between their elements. If you test any of them against the $(m, n)R(p, q)$ relation, the relation holds true: $$(2, 0)R(4, 2) \iff 2+2=0+4 \iff 4=4$$

Now, here's where the cleverness starts to come in: $$-1=\{(0, 1), (1, 2), (2, 3), ...\}$$ All of these elements have a difference of $-1$ between their elements. If you test any of them against the $(m, n)R(p, q)$ relation, the relation holds true: $$(1, 2)R(2, 3) \iff 1+3=2+2 \iff 4=4$$

Thus, using this partition, we were able to define a negative number using only addition and ordered pairs of natural numbers by making $(1, 0)$ maps to $+1$ while $(0, 1)$ maps to $-1$, which is the small trick behind all of this complex relation stuff.

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    $\begingroup$ you did mean "meaning no difference" right ? $\endgroup$ Commented May 25, 2016 at 23:52
  • $\begingroup$ @JoséOsorio Yes, that's right. I've fixed the typo above. Thanks for the catch! $\endgroup$ Commented May 25, 2016 at 23:54
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    $\begingroup$ I think $-1$ has $(0,1)$ not $(1,0)$ in its class. $\endgroup$
    – Ian
    Commented May 25, 2016 at 23:55
  • $\begingroup$ @Ian Also fixed. Thanks! $\endgroup$ Commented May 25, 2016 at 23:56
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    $\begingroup$ It might be helpful to note that this is basically exactly how the rational numbers are defined when you meet them as a child, but (1) formalized and (2) with subtraction instead of division. $\endgroup$ Commented May 26, 2016 at 10:59
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Formally you can't define a subtraction on $\mathbb{N}$, since an expression like $2-3$ wouldn't make any sense in $\mathbb{N}$. Nevertheless you can still define the subtraction implicitly via the addition by rewriting the relation $$b-x = a \Leftrightarrow a+x = b.$$ With the equation on the right-hand side $x$ is uniquely determined by the pair $(a,b) \in \mathbb{N} \times \mathbb{N}$, but choosing a different pair $(a',b')$ could result in the same $x$ as before, for example for $(a',b')=(a+1,b+1)$ we get $$a+x = b \Leftrightarrow (a+1)+x = (b+1).$$ You now want to identify all pairs $(a,b)$, which define the same $x$, so your goal is $$(a,b) \sim (a',b') :\Leftrightarrow (a,b) \text{ and } (a',b') \text{ define the same } x.$$ This is exactly done by the equivalence relation you mentioned above and the set of all solutions to the problem $a+x=b$ for $a,b \in \mathbb{N}$ is obviously $\mathbb{Z}$.

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  • $\begingroup$ You said formally i cna't define subtraction, and i been thinking about this since i read it, what if i gave a condition that made a-b>1 for a and b in naturals?, would someone ( i mean something) stop me from doing that? $\endgroup$ Commented May 26, 2016 at 0:59
  • $\begingroup$ In mathematics I guess nothing can stop you from doing what you want. xD But what you obtain if you do that is not a total function anymore, let me make that more precise: The addition +(a,b) := a+b is a binary operator (or a function with two arguments if you want) which maps NxN into N. If you would define the subtraction -(a,b) := a-b only for these numbers where a-b > 0, then it's not defined on the whole NxN anymore but on a proper subset of NxN and is a so called partial operator (or partial function). So yeah, in that sense you can define a subtraction on N (NxN to be precise). $\endgroup$ Commented May 26, 2016 at 10:16

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